1 Introduction to Differential Equations


 Dwain Gilmore
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1 1 Introduction to Differential Equations A differential equation is an equation that involves the derivative of some unknown function. For example, consider the equation f (x) = 4x 3. (1) This equation tells us information about the derivative f (x) of some function f(x), but it doesn t actually give us a formula for f(x). Of course, in this example, it s not too hard to figure out what f(x) might be. If the derivative of f(x) is 4x 3, then one possibility is that f(x) = x 4. This formula is a solution to the differential equation, because it matches the information about f (x) that we were given. Most differential equations have more than one solution. For example, f(x) = x and f(x) = x are also solutions to equation (1), since the derivative of either of these is equal to 4x 3. More generally, any formula of the form Really all we ve done here is integrate, i.e. 4x 3 dx = x 4 + C. f(x) = x 4 + C (2) is a solution, where C can be any constant, and every possible solution has this form. Thus, this formula is the general solution to equation (1). The general solution to a differential equation usually involves one or more arbitrary constants. Because of this, most differential equations have infinitely many different solutions, with one solution for every possible value of the constant(s). Letters other than y are often used in applications. Notation When writing differential equations, it is common to use the letter y for the function, instead of the letter f. Thus, equation (1) would be written y (x) = 4x 3. It is common in applications to use the variable t (for time) in place of x. In this case, the derivative would be written y or dy/dt, or possibly ẏ (with a dot above the y). Moreover, it is common to simply write y or dy/dx instead of y (x): y = 4x 3 or dy dx = 4x3. In any case, the goal is to find a formula for y in terms of x that satisfies the given differential equation. Basic Examples Let s look at a few more examples of differential equations, to help us get a feel for the subject.
2 INTRODUCTION TO DIFFERENTIAL EQUATIONS 2 EXAMPLE 1 Find the general solution to the following differential equation. y = 2x cos ( x 2). SOLUTION integrate: This isn t much harder than our initial example. All we need to do is 2x cos ( x 2) dx. This integral is most easily evaluated by substituting u = x 2, which gives cos u du = sin u + C = sin ( x 2) + C In general, any differential equation of the form y = f(x) can be solved by integrating: f(x) dx. However, not every differential equation is so simple. EXAMPLE 2 Find the general solution to the following differential equation. y = y. Actually, it is essentially the definition of the number e that the derivative of e x is e x. That is, e is defined so that e x is a solution to the differential equation y = y. SOLUTION This equation is much more interesting than those we have encountered so far. In words, it says the following: find a formula that doesn t change when you take its derivative. Here is one solution that immediately leaps to mind: e x. The derivative of e x is just e x, so if e x then y will be equal to y. But how can we find the general solution? Where should we place the constant C? So far, we have always just added C to the end of the formula: e x + C But this doesn t work: if e x + C, then y is just e x, so y and y are not the same. The answer is that C should be the coefficient instead. Ce x This works, since the derivative of Ce x is Ce x for any constant C. In the last example, we started by finding a particular solution to the dfferential equation, and then we figured out how to add a constant C to get the general solution. This is a common twostep process when solving differential equations EXAMPLE 3 Find a general solution to the following differential equation. y = y 2
3 INTRODUCTION TO DIFFERENTIAL EQUATIONS 3 SOLUTION Where do we even begin? How can we possibly find a formula for y that satisfies this equation? When we have no idea how to solve a math problem, one method we can always resort to is guess and check. We have no idea what formula might work here, and there s no way to find out without making some guesses, and then checking whether they work. Here are some possible guesses: sin x, e x, x, 1 x, x2, ln x. Do any of these work? It turns out that 1/x is the right guess. If 1/x, then y = 1/x 2, and the equation becomes 1 x 2 = ( 1 x) 2. Later on, we will learn a method called separation of variables that allows us to solve this equation without any guessing. So 1/x is a particular solution to this differential equation. Now, what about the general solution? We need to figure out how to include an arbitrary constant C. Here are a few possibilities: 1 x + C, C x, 1 Cx, 1 x + C, 1 x C. Actually, this answer isn t quite correct. In particular, 0 is a solution to the given equation, though it doesn t correspond to any value of C. (Intuitively, it s the solution you get when C =.) Do any of these work? Yes it is easy to check that 1 x + C is always a solution, so presumably this is the general solution to the given differential equation. Making up Differential Equations Although our goal is to learn how to solve differential equations, you can learn a lot by trying to make up differential equations that have a certain solution. For example, suppose we want a differential equation that has as a solution. The simplest possibility is x 3 y = 3x 2. However, any differential equation that holds when you plug in x 3 and y = 3x 2 will work. For example, xy = 3y has x 3 as a solution, since x ( 3x 2) = 3 ( x 3). Some other differential equations with x 3 as a solution include ( y )3 = 27y 2, xy + 4 7x 3, and yy = 3x 5. On your own, you could try making some differential equations that have x 2 solution, or perhaps sin x. as a
4 INTRODUCTION TO DIFFERENTIAL EQUATIONS 4 Initial Value Problems An initital value problem consists of the following information: 1. A differential equation involving an unknown function y. Instead of including y(0), sometimes an initial value problem includes a different value of y, such as y(1), or possibly the limiting value of y as x or x. 2. An initial value for y, i.e. the value y(0) of the function y when x = 0. The idea is that the value of y(0) is usually enough information to specify a single solution to the differential equation. Usually, an initial value problem has only one solution. Here is the most common procedure for solving an initial value problem. SOLVING INITIALVALUE PROBLEMS 1. Find the general solution to the given differential equation, involving an arbitrary constant C. 2. Plug in the initial value to get an equation involving C, and then solve for C. EXAMPLE 4 Find the solution to the following initial value problem. y = y 2, y(0) = 3. SOLUTION We found the general solution to this differential equation in Example 3: 1 x + C Therefore, all that remains is figure out a value for C so that y(0) = 3. Plugging in x = 0 and 3 gives the equation 3 = Solving for C gives C = 1/3, and hence C. 1 x + 1/3 We can simplify our answer by multiplying the numerator and denominator by 3: 3 3x + 1 EXAMPLE 5 Find the solution to the following initial value problem. y = 2y, y(0) = 5. SOLUTION The given differential equation isn t very different from the equation y = y from Example 2. In that case, the general solution was Ce x. How can we modify this solution to account for the extra 2?
5 INTRODUCTION TO DIFFERENTIAL EQUATIONS 5 A few minutes of thought reveals the answer: More generally, the solution to any equation of the form y = ky (where k is a constant) is Ce kx. Ce 2x So this is the general solution to the given equation. Plugging in x = 0 and 5 gives the equation 5 = Ce 0, so C = 5, and the solution is 5e 2x. When using t (for time) instead of x, the second derivative is sometimes written with two dots, i.e. ÿ. SecondOrder Equations Recall that the second derivative of a function y is the derivative of the derivative. This can be written y d 2 y or dx 2. A secondorder equation is a differential equation that involves y, as well as perhaps y, y, and x. EXAMPLE 6 Find the general solution to the following secondorder equation. y = 12x 2. SOLUTION Integrating once gives a formula for y : y = 12x 2 dx = 4x 3 + C. We can now integrate again to get a formula for y. (4x 3 + C ) dx = x 4 + Cx + C 2. Here C 2 represents a new constant of integration, which may be different from the original C. Actually, it would make more sense to refer to the original C as C 1 : x 4 + C 1 x + C 2 This is the general solution to the given secondorder equation. Note that the general solution in the last example involved two arbitrary constants. This is fairly common. 1. The general solution to a firstorder equation usually involves one arbitrary constant. 2. The general solution to a secondorder equation usually involves two arbitrary constants. Here the phrase firstorder equation refers to an equation that has only first derivatives, i.e. the sort of equation we were discussing initially. Incidentally, it is of course possible to discuss thirdorder equations (involving the third derivative), fourthorder equations, and so forth. As you would expect, the general solution to an nth order equation usually involves n arbitrary constants. However, we will mostly restrict our attention to first and second order equations, since equations of third order or higher are rare in both science and mathematics.
6 INTRODUCTION TO DIFFERENTIAL EQUATIONS 6 EXAMPLE 7 Find the general solution to the following secondorder equation. y = y. SOLUTION Obviously e x is a solution, and more generally C 1 e x is a solution for any constant C 1. However, this is not the general solution we are expecting one more arbitrary constant. So how can we find another solution to this differential equation? Think about this for a minute we want a function other than a multiple of e x that is equal to its own second derivative. The answer is quite clever: what about e x? Though the derivative of e x has an extra minus sign, the second derivative is again e x, so e x is a solution to the above equation. Indeed, anything of the form C 2 e x is a solution, where C 2 can be any constant. But how can we combine the two solutions into a single formula? In this case, it turns out that it works to just add them together: C 1 e x + C 2 e x (The reader may want to check this by plugging this formula into the original equation.) This formula includes two arbitrary constants, so it ought to be the general solution to the given secondorder equation. It is common in applications that the two known values of y are at the boundary points of the interval of possible xvalues. Hence the terminology boundary value problem. Because the general solution to a secondorder equation involves two arbitrary constants, you need two additional pieces of information to determine a single solution. One option is to give two different values for y, e.g. y(0) and y(1). This is called a boundary value problem, and you can solve it using the following procedure. SOLVING BOUNDARYVALUE PROBLEMS 1. Find the general solution to the given secondorder equation, involving constants C 1 and C Plug in the first value for y to get an equation involving C 1 and C Plug in the second value for y to get another equation involving C 1 and C Solve the two equations for the unknown constants C 1 and C 2. EXAMPLE 8 Find the solution to the following boundaryvalue problem y = 12x, y( 1) = 3, y(1) = 5. SOLUTION We can integrate to get a formula for y : y = 12x dx = 6x 2 + C 1, and then integrate again to get a formula for y: (6x 2 + C 1 ) dx = 2x 3 + C 1 x + C 2, All that remains is to find the values of C 1 and C 2. Plugging in x = 1 and 3 gives the equation 3 = 2 C 1 + C 2,
7 INTRODUCTION TO DIFFERENTIAL EQUATIONS 7 and plugging in x = 1 and 5 gives the equation 5 = 2 + C 1 + C 2, We can solve these two equations to get C 1 = 1 and C 2 = 4, so 2x 3 x + 4 Instead of giving two pieces of information about y, another way of specifying a single solution to a secondorder differential equation is to give one piece of information about y and one piece of information about y. In particular, a secondorder initial value problem consists of the following information: 1. A secondorder differential equation involving an unknown function y. 2. An initial value for y, such as y(0). 3. An initial value for y, such as y (0). You can solve such a problem using the following procedure. SOLVING SECONDORDER INITIAL VALUE PROBLEMS 1. Find the general solution to the given secondorder equation, involving constants C 1 and C Plug in the initial value for y to get an equation involving C 1 and C Take the derivative of the general formula for y to get a general formula for y. 4. Plug in the initial value for y to get another equation involving C 1 and C Solve the two equations for the unknown constants C 1 and C 2. EXAMPLE 9 Find the solution to the following initial value problem. y = y, y(0) = 7, y (0) = 3. SOLUTION As we saw in Example 7, the general solution to the given equation is C 1 e x + C 2 e x. Therefore, we need only figure out the values of C 1 and C 2. Plugging in x = 0 and 7 gives the equation 7 = C 1 + C 2. Next we take the derivative of the general formula for y to get a general formula for y. y = C 1 e x C 2 e x. Plugging in x = 0 and y = 3 gives the equation 3 = C 1 C 2. We can now solve the equations C 1 + C 2 = 7 and C 1 C 2 = 3 for C 1 and C 2. The result is that C 1 = 5 and C 2 = 2, so 5e x + 2e x
8 INTRODUCTION TO DIFFERENTIAL EQUATIONS 8 Guessing the Form One basic method for solving differential equations is an enhanced version of guess & check: we can guess the form of the solution, and then solve for any missing constants. For example, we might guess that a differential equation has a solution of the form e ax for some unknown value of a. We then check this solution by plugging it into the differential equation, and then try to figure out which values of a will make the solution work. EXAMPLE 10 Suppose we wish to find a solution the equation y = 7y 10y. We might guess that this equation has solutions of the form e ax for some constant a. In this case, we have y = ae ax and y = a 2 e ax. Plugging these into the equation gives which simplifies to a 2 e ax = 7ae ax 10e ax a 2 e ax = (7a 10)e ax. Now, how can we arrange it so that the left and right sides of this equation are the same? Well, they will be the same as long as For this equation we only found two solutions e 2x and e 5x of the given form, but there ought to be many more solutions not of this form. Indeed, the general solution to this differential equation is C 1 e 2x + C 2 e 5x. a 2 = 7a 10 The solutions to this quadratic equation are a = 2 and a = 5, and therefore e 2x and e 5x are two solutions to the given differential equation. EXAMPLE 11 Find a solution to the equation x 2 y = 2xy + 10y of the form x a. SOLUTION If x a, then y = ax a 1 and y = a(a 1)x a 2 Plugging these into differential equation gives which simplifies to x 2( a(a 1)x a 2) = 2x ( ax a 1) + 10x a a(a 1)x a = (2a + 10)x a. Now, how can we arrange it so that the left and right sides of this equation are the same? Well, they will be the same as long as Again, we only found two solutions to the given equation, but it turns out that the general solution is C 1 x 5 + C 2 x 2. a(a 1) = 2a + 10 Solving gives a = 5 or a = 2, so x 5 and x 2 are two solutions to this equation.
9 INTRODUCTION TO DIFFERENTIAL EQUATIONS 9 Forms with Two Constants Sometimes it works well to guess a form that involves two constants. For example, consider the equation y y = 2y We might guess that this equation has solutions of the form ax b, where a and b are constants. The derivatives of this form are y = abx b 1 and y = ab(b 1)x b 2. Plugging these into the differential equation and simplifying yields a 2 b 2 (b 1)x 2b 3 = 2ax b. The only way for the left and right sides of this equation to be the same is if the coefficients are the same and the exponents are the same. This gives us the following two equations: a 2 b 2 (b 1) = 2a and 2b 3 = b. The second equation tells us that b = 3. Plugging this into the first equation and solving for a yields a = 0 or a = 1/9. Therefore, 0 and 1 9 x3 are two solutions to the given equation. EXERCISES 1 6 Use integration to find the general solution to the given differential equation. 1. y = x x y = x cos x 3. y + cos(3x) = 0 4. y e x = 1 5. xy + 4x 3 = 1 6. y = 1 x 2 y 7. y = 3 x 8. x 3 y = x Use guess & check to find the general solution to the given differential equation. 9. y + y tan x = ( y )2 = 4y Use guess & check to find just one solution to the given differential equation. 11. y + 9e 2x 12. yy = 4e 8x 13. x 2 y + e 2x 14. y y = 14y + 4x Solve the given initial value problem. 15. y = xe x, y(0) = y = 3y, y(2) = Solve the given boundary value problem. 17. y = sin x, y(0) = 4, y(π) = y = y, y(0) = 7, y(ln 2) = Solve the given initial value problem. 19. y = x 2, y(1) = 1/2, y (1) = 1/2 20. y = 4y, y(0) = 5, y (0) = Find all solutions to the given differential equation of the specified form. 21. y = 3y + 4y ( e ax ) 22. x 2 y 7xy ( x a ) 23. y y = 36y 3 ( x a ) 24. 5yy = ( y )2 + 36y 2 ( e ax )
10 Answers ( x ) 3/2 +C 2. x sin x + cos x +C sin(3x) +C 4. e x +C 5. ln x 4 3 x3 +C 6. arctanx +C x7/3 +C 1 x +C 2 8. ln x + x 1 +C 1 x +C 2 9. C cos x 10. (x +C) e 2x 12. e 4x 13. lnx 14. x (x 1)e x e 3x sinx + (2/π)x e x + 4e x ( x 4 + 2x + 3 ) 20. 3e 2x + 2e 2x 21. e x and e 4x 22. x 2 and x x e 3x and e 3x
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